Chapter 5 : Unit-root Testing and Cointegration Analysis

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Chapter 5 : Unit-root Testing and Cointegration Analysis

5.1 Introduction to spurious regression

• Suppose that we observe the following regression results(t statistics in parentheses):

Yt = 449.95 + 14.794 X1t – 0.71501 X2t + 0.4393 X3t + t

(1.474) (5.131) (– 9.528) (13.903)

n = 1952 – 1988 ; R2 = 0.9253 ; DW = 0.4127

Aside from some concerns with the relatively low DW statistic, you should be very pleased with the results: high R2, high t ratios, etc.,

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But note :

Yt : Population in New Zealand;

X1t : Price level in China;

X2t : Money supply in China;

X3t : Per Capita Income in China.

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Consider another amazing example

(from Hendry (1980): Econometrica):

Yt = 10.9 – 3.2 X1t + 0.39 X1t2

(19.82) (13.91) (19.5)

n = 1945 – 1970 ; R2 = 0.982 ; DW = 0.1

Notes :

Yt = Logarithms of C.P.I. in the U.K.;

X1t = cumulative rainfall in the U.K.

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Obviously economic variables of China have no influence on the population of New Zealand. It would be equally absurd to conclude that the wet weather in Britain explains the rapid inflation experienced by the British economy. These are examples of “spurious regression”. Results occur because we are working with non-stationary data.

When using time series data, usual t-test; properties of R2 and OLS estimators, only typically applicable when variables are stationary. Signs of non-stationary are low DW but high R2.

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5.2 Stationarity and Integration

Let yt be a time series variable. Then yt is stationary if the process which generates yt is invariant with respect to time:

If yt is stationary, then

(i) its mean E (yt);

(ii) its variance E((yt – E(yt))2);

(iii) covariance for any lag k; Cov(yt , yt-k)

are also invariant with respect to time.

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AN EXAMPLE OF A STATIONARY PROCESSY = 0.7*lag(Y) + e

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AN EXAMPLE OF A NON-STATIONARY PROCESSY = 1*lag(Y) + e

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Homogenous Non-stationary Process

Very few economic time series are stationary. But we can often “difference” a series to form a stationary process. Such a non-stationary series is called homogenous.

The order of homogeneity or order of integration is the number of times needed to difference the series to obtain a stationary series. Given this, we define a stationary series is defined as “integrated of order 0”, or I(0)

Specifically, we say

Yt is I(d)

That is, Yt is integrated of order d if

d Yt is I(0),

where is the difference operator and d is the number of differences.

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Example : Suppose Yt is a non-stationary homogenous series. Then if

((( Yt = Yt – Yt-1 is stationary. That is, Yt ~ I(0).

Then Yt is integrated with order 1.

(b) Yt is also non-stationary, but 2 Yt ~ I(0), then Yt ~ I(2).Generally, most economic time series are I(1) or I(2). Rarely are they I(0).

So what? Serious implications for regression analysis. Classical results, as discussed so far, assume data series are all stationary. Distribution of conventional statistics and estimators (for example, t ratios; regression coefficients; R2; F Ratio) for regression involving non-stationary variables are (typically) not at all like those derived under stationarity. So, it is very important to test for stationarity prior to proceeding with econometric analysis, if we want to avoid spurious results.

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Now, suppose Yt ~ I(1), then

Yt could follow a simple random walk process, That is,

Yt = Yt-1 + t

t is white noise. That is, t is an independently distributed random variable with zero mean and variance 2. In other words, Yt = Yt – Yt-1 is a white noise process. Examples of random walk series are stock prices; future contracts; interest rates; exchange rates.

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There are substantial differences between an I(0) series and an

I(1) series such as random walk.

(a) I(0) series has a finite mean and variance. There is a tendency for the the series to return to the mean. Conversely, a random walk will wander widely and will rarely return to an earlier value. Random walk has an infinite variance.

(b) Autocorrelation functions and Partial Autocorrelation Functions of I(0) and I(1) are different.

Autocorrelation for an I(0) series declines rapidly as lag(k) increases

The process gives low weight to events in the medium to distant past. That is,

I(0) series has a finite memory.

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The Autocorrelation Function for an I(1) series are all near 1 in magnitude, even for large k.

The process has indefinitely long memory.

Similar pictures for the Partial Autocorrelation functions. For stationary series should decline to zero, while high magnitudes for non-stationary series.

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Whether an economic time series is I(0) or I(d) , d 0, should be of serious concern to policy makers as it indicates whether the implication of a policy shock (tax change, welfare change etc.) are temporary or permanent.

For an I(0) series, any shocks are temporary. There is always a tendency for the series to move towards its finite. Constant mean. For an I(1) series, shocks have permanent effects.

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General form of random walk

(a) Random walk with drift

Yt = + Yt-1 + t

accounts for the non-zero mean in the differenced series.

(b) Random walk with drift and trend:

Yt = + Yt-1 + t + t

t = 1, 2, 3, …

That is,

Yt = + t + t

So, general form of an I(1) series is

Yt = Yt-1 + + t + t (1)

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Remarks:

(i) (1) is called a “unit root” model as it is a special case of a geneal AR(1) model :

Yt = Yt-1 + +t + t

with = 1

(ii) If = 1, OLS estimation of (1) is inappropriate as data are non-stationary. Estimators and test statistics do not have the usual properties.

(iii)Removing drift and time trend does not help, as we are still left with the stochastic trend:

Yt = Yt-1 + t

Testing for a unit root ( = 1) was developed in the literature only in the last twenty years or so. A very popular test is the Dickey-Fuller (DF) (1981) test.

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5.3 Testing for a unit root or the order of integration

Suppose

Yt = Yt or Yt-1 + +t + t 1 (a)

Then recall:

(i) Random walk or unit root if = 1. Yt or detrended Yt are both non-

stationary. OLS is not appropriate.

(ii) If 1 and > 0, then Yt growing because of positive deterministic

trend.

Detrended Yt is I(0). OLS is appropriate.

So, we test H0 : = 1. Unfortunately we cannot estimate (a) by OLS and do t-test using . OLS is biased towards zero if = 1 and so we could incorrectly reject H0 : = 1.

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Dickey-Fuller (DF) test :

From (a)

Yt = Yt-1 + +t + t

Yt = ( – 1) Yt-1 + +t + t

= Yt-1 + +t + t

If Yt is non-stationary, then = 1 and = 0

If Yt is stationary, then < 1 and < 0

So, testing = 1 is equivalent to testing

H0 : = 0 vs H1 : < 0

If we do not reject H0, we believe Yt is I(1).

If we reject H0, we believe Yt is I(0).

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To undertake the DF test:

Estimate Yt = Yt-1 + +t + t by OLS. Obtainand the associated t-statistics.Dickey and Fuller derived distribution of t-statistic: .Test : < 0 compare t-stat with critical values from Table 1.

Remarks :1. Dickey and Fuller derived large sample distribution of t = . Note that is not Student’s t distributed.

Critical values for the test were calculated via Monte-Carlo simulation methods. Different authors report different critical values for the same sample size. We will use Dickey-Fuller’s tables.

ˆ,ˆ,ˆ

ˆ/ˆ se

ˆ/ˆ se ˆ/ˆ se

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10 vs.0: HH

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2. Also, we need different critical values if we estimate(a) Yt = Yt-1 + t

(b) Yt = + Yt-1 + t

3. However, it is usual to include a drift term , as = 0 implies the change series has zero mean which is not typically the case for economic data.

4. What about the linear time trend? No agreed strategy, but one approach (proposed by Dolado, Jenkinson and Sosvilla-Rivero (1990)) is to :

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(a) Estimate the following equation by OLS :

Yt = Yt-1 + + t + t

and obtain .

(b) Test < 0 using “t-test”. Compare test statistic with critical values from Table 1.

(c) If we reject H01, then we believe Yt is stationary.

(d) If we cannot reject H01, then test

using a F type test. Compare calculated test statistic with Table 2.

(e) If we reject H02 test H0

1 again using the standard normal tables. If we reject H0

1, then Yt is stationary.

ˆ,ˆ,ˆ11

10 vs.0: HH

0;0:vs.0: 21

20 HH

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(f) If cannot reject H02, then estimate

Yt = + Yt-1 + t

(That is, no trend term)

and test

H03 : = 0 vs H1

3 : < 0

using “t-test” and compare with critical values in table 3.

If we reject H03 then we believe that the series is stationar

y. If we cannot reject H03, then the series is non-stationa

ry. (We could also test but is usually significant with economic data).

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Example 1 :

Suppose we estimate

Yt = + Yt-1 + t + t ; T = 50

Test H01 : = 0 vs H1

1 : < 0 using “t-test”

(i) Suppose the calculated test statistic is –4.20. From Table 1, 10% critical value is – 3.18. So, we reject H0 and Yt is believed to be stationary. No further testing is required.

(ii) Suppose instead the calculated test statistic is –2.50. We cannot reject H0. So now test H0

2 : = = 0 vs H12 : < 0 ;

0 using a F type test. Suppose the calculated test statistic is 9.32. 10% critical value from Table 2 is 5.61. We reject H0

2 .

Then, re-test H01 : = 0 vs H1

1 : < 0 using standard normal

10% critical value = – 1.2816. Clearly, –2.50 < –1.2816. We reject H0

1 and believe that Y1 is stationary.

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For these test to be valid, we must have white noise residuals in the regression.

Frequently this is not the case. So it leads to the Augmented Dickey Fuller (ADF) test.

The aim is to add sufficient lags of Yt-j to ensure no autocorrelation in the residuals. But how to find P?

One way is to find the AF and PAF for the residuals from estimating (*) with P = 0, 1, 2 … until residuals are white noise. Then, having determined P, follow the testing strategy as given for DF test.

t

P

jjtjtt YYtY

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Testing for a unit root in the New Zealand population series using the Dickey-Fuller test

In this example, our aim is to determine the order of integration for the New Zealand population, over 1890-1991. The series in levels is depicted below – it is clearly non-stationary – at least in its mean. POP is also graphed – it is difficult to “guess” whether POP is stationary – its mean could well be, but the mean is clearly non-zero, suggesting that the D.F. models will require a drift term. There may well also be a positive trend but this is extremely difficult to ascertain from the graph.

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NZ POPULATION (1890-1991)

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ANNUAL CHANGE IN NZ POPULATION 1891-1991

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We now turn our attention to formally testing, using the DF test, weather POP has a unit root – i.e., is the series stationary. Recall that the general model under test is :

POPt = POPt-1 + + t + t (1)

where, in particular, it is assumed that the disturbance is white noise. Recall from our earlier discussion that we first estimate (1) by OLS and test

H0 : = 0 vs H1 : < 0 (2)

using the “t-test” and the critical value from Table 1. The SAS commands and output follow:

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data nzpop;input pop;t =_n_;lpop-lag(pop);dpop=pop-lpop;proc reg;model dpop=lpop t;run;

Analysis of Variance

Source DF Sum of Squares Mean Square F Value Prob>F

Model 2 3833.14297 1916.57148 6.695 0.0019Error 98 28056.23486 286.28811C Total 100 31889.37782

Root MSE 16.92005 R-square 0.1202

Dep Mean 26.82673 Adj R-sq 0.1022

C.V. 63.07160

Parameter Estimates

Variable DF Parameter Estimate Standard ErrorT for H0 :

Parameter=0 Prob > |T|

INTERCEP 1 24.752368 5.25672933 4.709 0.0001LPOP

1 −0.020965 0.01127235 −1.860 0.0659T 1 0.786373 0.33006833 2.382 0.0191

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The appropriate t-stat is – 1.860. From Table 1, the 10% critical value for 100 observations (we have 101 here) is 1 3.17. Clearly, as expected, we cannot reject H0. Given this we now need to test:

H0 : = = 0 vs H1 : < 0, 0 (3)

using the “F-test”. This is usually undertaken using the “Lagrange Multiplier” or “Likelihood Ratio” tests.

test t = 0, lpop=0;

Dependent Variable: DPOP

Test: Numerator: 1916.715 DF: 2 F value: 6.6946

Denominator: 286.2881 DF: 98 Prob>F: 0.0019

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The calculated test statistic is 6.695. From Table 2 the 10% significance level critical value is 5.47 (n = 100). So we reject H0 , and believe that the trend is significant. The procedure is to now re-test hypothesis (1) using the standard normal tables – at the 10% level and the critical value is –1.2816; –1.6449 at the 5% level, and –2.3263 at the 1% level. Our test statistic is –1.85988, so we reject H0 at the 5% level but not at the 1% level. Given this I would err on the conservative side and proceed as if we cannot reject H0 – that is, proceed as if POP is non-stationary and at least I(1).

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Why at least I(1)? We have merely rejected that the series is stationary, it could be I(1) or I(2) etc. Our next step is to rerun the experiment using second difference, that is, estimate :

2 POPt = POPt-1 + + t + t (4)

Let POPt = DPOP then we can write (4) as :

DPOPt = POPt-1 + + t + t (5)

Which is exactly the same form as (1) except in terms of DPOP. So, our interest now lies in whether = 0 in (5). If we cannot reject that = 0 then we believe that DPOP is I(1), i.e., POP is I(2). Conversely, if we reject the hypothesis that = 0, then we believe that DPOP is I(0), i.e., POP is I(1). If this occurs and given our previous result that POP is not I(0) then POP must be I(1).

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Following are the SAS commands and output for estimating (4), or equivalently (5).data nzpop2;set nzpop2;1dpop=lag(dpop);d2pop=dpop-1dpop;proc reg;model d2pop=1dpop t;run;

Model: MODEL1Dependent Variable: D2POP Analysis of Variance



Root MSE 1187544 R-square 0.1403Dep Mean 0.24100 Adj R-sq 0.1226C.V. 4927.56647

Parameter Estimates



INTERCEP 1 5.650266 2.73540305 2.066 0.0415LPOP 1 –0.276650 0.06976169 –3.966 0.0001ONE 1 0.038279 0.04315646 0.887 0.3773

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The t-stat for the hypothesis of H0 : = 0 vs. H1 : < 0 is –3.966. From table 1, the 10% critical value is–3.17 (n = 100). We reject H0 – DPOP is I(0), so POP is I(1). We identify that the order of integration for POP is 1. Note that most economic time series are I(1) or I(2) – if a series is I(2) then we would accept H0 : I(2) vs. H1 : I(1) and so we take another difference and test H0 : I(3) vs. H1 : I(2), and so until we believe that H1 is appropriate.

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Testing for a unit root in a New Zealand population series using the Augmented Dickey-Fuller test

We first need to determine the order of augmentation (if any) that is required to ensure that the residuals from the integrating regression are (approximately) white noise. That is, what is the value of p in

Below are the commands and outputs for determining p.

t

p

jjtjtt POPtPOPPOP

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data nzpop;infile ‘c:\teaching\ar3316\nzpop.dat’;input pop;t=_n_;lpop=lag(pop);dpop=pop-lopo;proc reg;model dpop=lpop t;output out=out1 r=e;run;proc arima data = out1;identify var=e nlag=10;run;

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Model: MODEL1Dependent Variable: DPOP





Dep Mean 26.82673 Adj R-sq 0.1022

C.V. 63.07160

Parameter Estimates



INTERCEP1 24.752368 5.25672933 4.709 0.0001

LPOP1 −0.020965 0.01127235 −1.860 0.0659

T1 0.786373 0.33006833 2.382 0.0191

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ARIMA Procedure

Name of variable = E .

Mean of working series = 1.53E-14

Standard deviation = 16.66687

Number of observations = 101

Autocorrelations

Lag Covariance Correlation −1 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 1

0 277.785 1.00000 . ********************

1 200.032 0.72010 . **************

2 125.992 0.45356 . *********

3 72.937728 0.26257 . ***** .

4 50.922254 0.18332 . **** .

5 30.312565 0.10912 . ** .

6 38.052909 0.13699 . *** .

7 49.768591 0.17916 . **** .

8 63.723119 0.22940 . ***** .

9 60.939741 0.21938 . **** .

10 50.070769 0.18025 . **** .

marks two standard errors ״.״

Partial Autocorrelations

Lag Correlation −1 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 1

1 0.72010 . **************

2 −0.13496 .*** .

3 −0.02313 . .

4 0.08221 . ** .

5 −0.06576 . * .

6 0.16945 . *** .

7 0.05493 . * .

8 0.07872 . ** .

9 −0.00616 . .

10 −0.02255 . .

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Are the residuals from this regression white noise? If so, then p = 0; that is the degree of augmentation is zero and the DF test is appropriate. If not, then the DF test is inappropriate and we do need to augment the integrating regression

Clearly, there is significant autocorrelations at k = 1 and k = 2 – the residuals are not approximately white noise, and the DF test is not appropriate. We now ask whether the residuals from the integrating regression with p = 1 are uncorrelated.

data nzpop2;set nzpop;ldpop=lag(dpop);proc reg;model dpop=lpop t ldpop;output out=out2 r=e;proc arima data = out2;identify var=e nlag=10;run;

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Model: MODEL1Dependent Variable: DPOP





Dep Mean 27.05800 Adj R-sq 0.5642

C.V. 43.41524

Parameter Estimates



INTERCEP 1 10.612394 3.89798387 2.723 0.0077LPOP 1 −0.014089 0.00796622 −1.769 0.0801T 1 0.448239 0.23570633 1.902 0.0602LDPOP 1 0.714694 0.06918228 10.331 0.0001

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ARIMA Procedure


Mean of working series = 3.73E-15



Autocorrelations


0 132.479 1.00000 ********************

1 10.200563 0.07700 . ** .

2 −3.743495 − 0.02826 . * .

3 −15.294723 −0.11545 . ** .

4 3.766422 0.02843 . * .

5 −18.210605 −0.13746 . *** .

6 2.446422 0.01847 . .

7 3.585575 0.02707 . * .

8 18.020811 0.13603 . *** .

9 9.979367 0.07533 . ** .

10 5.013424 0.03784 . * .



Lag Correlation −1 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 1

1 0.07700 . ** .

2 −0.03439 . * .

3 −0.11134 . ** .

4 0.04593 . * .

5 −0.15263 . *** .

6 0.03292 . * .

7 0.02362 . .

8 0.10246 . ** .

9 −0.07882 . ** .

10 −0.01759 . .

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Of course, we can formally test whether any of these autocorrelations are significantly different from zero using a t-test, but we’ll simply “eyeball” the functions and see if there are any “large” spikes. I would believe not. This suggests that we can obtain (approximately) white noise residuals by setting p = 1. That is, we estimate,

POPt = POPt-1 + + t + 1POPt-1 + t (1)

and using this model to test for stationarity of the series. We now proceed to do this following exactly the same strategy as we discussed for the DF test but applied to model (1) rather than to the DF integrating regression which assumes that p = 0. Fortunately, the critical values for the tests do not change as p changes.

We first test

H0 : = 0 vs. H1 : < 0

using the estimated t-stat on . From our printout this is −1.769. The 10% critical value from Table 1 of your handout with n = 100 is −3.17 – we cannot reject H0. We then proceed to test whether or not the trend term is significant :

H0 : = = 0 vs. H1 : < 0; ≠ 0

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So, the F stat is 1.966. From Table 2, the 10% critical value is 5.47 (n=100) – we cannot reject H0. That is, we cannot reject the trend is insignificant. Note that this is not the conclusion that we reached when we incorrectly used the DF test. So we now estimate,

(2)

determing p in the same way before. The commands and output for this follows:

t

p

jjttt POPPOPPOP

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test t = 0, lpop=0;

Dependent Variable: DPOP

Test: Numerator: 271.2860 DF: 2 F value: 1.9659


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Proc reg data=nzpop;model dpop=lpop;output out=out3 r=e;proc arima data=out3;identify var=e nlag=10;run;

Model: MODEL1Dependent Variable: DPOP Analysis of Variance


Model1 2208.14408 2208.14408 7.365 0.0078

Error99 29681.23374 299.81044

C Total100 31889.37782


Dep Mean 26.82673 Adj R-sq 0.0598

C.V. 64.54396

Parameter Estimates



INTERCEP1 16.685476 4.11487497 4.055 0.0001

LPOP1 0.005477 0.00201820 2.714 0.0078

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ARIMA Procedure


Mean of working series = −3E-16



Autocorrelations


0 293.874 1.00000 ********************

1 213.018 0.72486 . **************

2 136.511 0.46452 . *********

3 82.185089 0.27966 . ******

4 59.896416 0.20382 . **** .

5 38.417678 0.13073 . *** .

6 45.903035 0.15620 . *** .

7 57.612109 0.19604 . **** .

8 71.602274 0.24365 . ***** .

9 68.460802 0.23296 . ***** .

10 56.732395 0.19305 . **** .



Lag Correlation −1 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 1

1 0.72486 . **************

2 −0.12833 . *** .

3 −0.01551 . .

4 0.08702 . ** .

5 −0.06306 . * .

6 0.17057 . *** .

7 0.05806 . * .

8 0.07958 . ** .

9 −0.00375 . .

10 −0.02382 . .

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Clearly, some significant autocorrelations. So, consider p=1,data nzpop3;set nzpop2;one=1;proc reg;model dpop=lpop ldpop one/noint;/output out=out4 r=e;proc arima data=out4;identify var=e nlag=10;

Model: MODEL1NOTE: No intercept in model. R-square is redefined.Dependent Variable: DPOP



Model 3 90815.76160 30271.92053 213.602 0.0001Error 97 13746.95840 141.72122U Total 100 104562.72000


Dep Mean 27.05800 Adj R-sq 0.8645C.V. 43.99687

Parameter Estimates

Variable DF Parameter Estimate Standard Error T for H0 : Parameter=0 Prob > |T|

LPOP 1 0.000810 0.00146216 0.554 0.5808LDPOP

1 0.731029 0.06956663 10.508 0.0001ONE 1 5.944300 3.06854422 1.937 0.0556

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ARIMA Procedure


Mean of working series = −143E-17



Autocorrelations


0 137.470 1.00000 ********************

1 11.350360 0.08257 . ** .

2 −3.645662 −0.02652 . * .

3 −16.040212 −0.11668 . ** .

4 2.955086 0.02150 . *** .

5 −19.830221 −0.14425 . .

6 1.625075 0.01182 . .

7 3.078413 0.02239 . .

8 18.188248 0.13231 . *** .

9 9.874910 0.07183 . * .

10 4.944657 0.03597 . * .



Lag Correlation −1 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 1

1 0.08257 . ** .

2 −0.03357 . * .

3 −0.11254 . ** .

4 0.04050 . * .

5 −0.15871 . *** .

6 0.02797 . * .

7 0.01829 . .

8 0.09760 . ** .

9 0.07192 . * .

10 0.01276 . .

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This looks reasonable – so we’ll proceed using p=1. We now test H0 : = 0 vs H1 : < 0 using the t-stat from this model with p=1. From the above printout this is 0.5541 – it is positive and so we know automatically that we cannot reject H0 . Following our strategy, given this result, we now need to test:

H0 : = = 0 vs. H1 : H0 not true

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We find the F statistic to be 5.708. the 10% critical value is 3.86 (from Table 4). So we reject H0 and believe that there is a significant drift term. We now need to re-test H0 : = 0 vs. H1 : < 0 using the standard normal critical value. Again, we know that these critical values are negative, and our t stat is 0.5541 – so we cannot reject H0 : POP is not I(0) and could be I(1).

We can continue this process and test I(1) vs. I(2) etc., and will eventually conclude that POP is I(1).

Test lpop=0, one=0;

Dependent Variable: DPOPTest: Numerator: 808.8862 DF: 2 F value: 5.7076


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49

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Integration results suggest that we should model non-stationary series when they are appropriately differenced. This is valid except when the variables of interest are non-stationary (integrated of the same order) and cointegrated. Then it is legitimite to estimate an economic model using undifferenced data.

What do we mean by variables being cointegrated? We will only consider the two variable case. Consider a pair of variables xt and yt, each of which is I(d). The linear combination of xt and yt

zt = yt – bxt (1)

is also I(d). However, if there exists a constant b such that zt is I(0), then xt and yt are cointegrated. b is called “cointegrating parameter”. For the two variable case b is unique.

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Remarks :

(a) (1) is called cointegrating regression.

(b) Recall that properties of I(0) & I(1) series’ are vastly different. So, if b is such that zt ~ I(0) then xt and yt must have a special relationship.

(c) Special relationship as represented by

yt = bxt (2)Is called the “long run” or “equilibrium” relationship, as suggested by economic theory.

(d) Given models (1) and (2), zt measures the extent to which the system is out of equilibrium. It is called the “equilibrium error”.

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(e) Here, “equilibrium” describes tendency of economic system to move towards a particular region of possible outcome space. If, for example, yt and xt are I(1) but “move together” in the long run, then zt must be I(0): yt and xt must be cointegrated. Otherwise, the series could drift apart without bound.

(f) Cointegrated variables may drift apart in the short run, but in the long run, economic forces such as market mechanism will bring them back to their equilibrium relationship.

Examples:

# Interest rates on assets on different maturities

# Prices of a commodity in different parts of a country

# Income and expenditure by local government

# Values of sales and production costs of a firm

# Maybe also : prices and wages, imports and exports, market prices of substitute commodities, money supply and prices, spot and future prices of a commodity.

53

(g) So, a test for cointegration is a test of the long run equilibrium relationship between the variables.

(h) If xt and yt are cointegrated, then b is the long run multipler. It has been shown that the OLS estimator of b is “super consistent” (converges to the true parameter at faster rate than usual).

(i) Obviously if xt and yt are cointegrated, then it is legitimate to model the relationship in non-stationary form.

54

Testing for cointegration

Numerous tests have been suggested to test for cointegration. We will consider only two:

1. CRDW (Cointegrating Regression Durbin-Watson) test

2. CRADF (Cointegrating Regression Augmented Dickey-Fuller) test

To illustrate we’ll assume that xt and yt are both I(1). To undertake tests, we estimate the cointegrating regression by OLS:

yt = + bxt + zt

(we can also include trend term)

The cointegrating residuals are

which should be I(0) if xt and yt are cointegrated. If not, then ttt xbyz ˆˆˆ (1). is ˆ Izt

55

CRDW test (Cointegrating Regression Durbin Watson test) :If has a unit root, then DW statistic for the cointegrating regression approaches zero. Why?Recall the case of AR(1) disturbance

DW test statistic aims to test

H0 : = 0 vs H1 : ≠ 0Now, recall that,

DW = 2(1 − )If there is a unit root then = 1, which implies that DW = 0.So, testing = 1 is equivalent to testing

H0 : DW = 0 ( is non-stationary xt and yt are not cointegrated)

H0 : DW > 0 ( xt and yt are cointegrated)

So, the test rejects H0 if CRDW is significantly greater than zero. We need to use different critical values. Suppose critical value is d*. Then if DW < d*, cannot reject H0; if DW > d*, reject H0 .

Remark :The CRDW test typically has low power. It is frequently regarded as a quick but not recessarily accurate test

tz

ttt zz 1ˆˆ

tz

56

CRADF test (Cointegrating Regression Augmented Dickey Fuller test) :

Apply an augmented DF test to

That is, we estimate

where again is selected to ensure t is white noise, in particular, t's are uncorrelated. Then follow testing strategy as per Augmented Dicky-Fuller stationary test.

But we have to use different critical values.

Why? To estimate (3), we have to first estimate the cointegrating regression. That is, we had to estimate the cointegrating parameter to form . It is found that the distribution of t statistic for H0 : = 0 vs. H1 : < 0 depends on the number of coefficients estimated in the cointegrating regression.

(For augmented Dicky-Fuller stationary test, we hadn’t estimated any coefficient prior to running test)

tz

tz

tjtj

jtt zzz

ˆˆˆ1

1

57

Remarks :

(i) CRADF test has relatively higher power than CRDW test

(ii) Choose p as in integration discussion. That is, we choose p such that the residuals from (3) have no significant auto correlations. Critical values depend on p. Table gives p = 0 and p = 4.

(iii) Asymptoically, it makes no difference whether we estimateyt = + bxt + zt

orxt = * + b*yt + z*

tbut in finite samples, it can matter. In practice, we tend to estimate both equations.

(iv) In the two variable case, we require that both variables to be integrated of the same order.

(v) Can extend notion of cointegration to multivariate caseyt = + b1x1t + b2x2t + … bkxkt + t

Problems become more complicated as cointegrating relationship need no longer be unique. Also, variable need not be integrated of the same order.

58

Error correction models (ECM)

Recall that in a cointegrating regression,

yt = + bxt + zt

and so,zt = yt – – bxt = error from equilibrium relationship.

If yt and xt are cointegrated, then this adjustment prevents zt from getting larger and larger. This had led to the so called Engle-Granger “Error Correction models (ECM)”. Such models aim to incorporate short run dynamics with long run equilibrium.

If yt and xt are both I(1) and are cointegrated, Engle and Granger (1987, Econometrica) show that there exists an “error correction” representation, which is free of spurious regression, of the form,

yt = 01 + 1zt–1 + lagged(yt , xt) + 1t

xt = 02 + 2zt–1 + lagged(yt , xt) + 2t

where zt = yt – – bxt is error term from cointegrating regression. 1t and 2t are jointly white noise, but possibly contemporaneously correlated.

Engle and Granger show that cointegrating variables must obey such a model. Further, data generated by ECM must be cointegrated.

59

Remarks :

(a) Asymptotically, it does not matter whether zt is obtained from yt = + bxt + zt or the reverse cointegrating regression. In finite samples, it can make a difference. Therefore, it is usual to define ECM as,

yt = 01 + 1zt–1 + lagged(yt , xt) + 1t

xt = 02 + 2zt–1 + lagged(yt , xt) + 2t

(b) ECM are not directly derived from economic theory. Estimated parameters only bear an indirect relationship to theoretical parameters of interest. Nevertheless, it is useful for forecasting and testing causality hypothesis.

(c) ECM’s can be extended to the multivariate case, yet detailed discussion is beyond the scope of this course.

(d) Note that ECM incorporates long run equilibrium via zt , and short run dynamics of adjustment process via the lag terms.

(e) If xt and yt are not cointegrated, then equilibrium concept has no practical implications. We should exclude the error correction term and simply write the model as,

yt = 01 + lagged(yt , xt) + 1t

xt = 02 + lagged(yt , xt) + 2t

This is usually called a Vector Autoregressive (V.A.R.) model

Conversely, if xt and yt are cointegrated, then omitting the error correction term results in mis-specification bias.

60

(f) If 1t and 2t are contemporaneously correlated, then we should estimate the ECM as a set of seemingly unrelated regression equations (S.U.R.E.).

(g) So the procedure to follow (commonly called the Engle-Granger two-step procedure) is

(i) Estimate the cointegrating regression and obtain . Determine if xt are cointegrated.

(ii) If xt and yt are cointegrated, then estimated the ECM as,

(iii) Otherwise, estimate the models as,

(h) There are lots of suggestions as to how to determine the appropriate number of lags to include. One such criterion is Akaike’s Final Prediction Error (FPE) in a two-step fashion. For example, in the first equation, we set q = 0 and vary m to find m = m* which minimizes

With m* chosen, we vary q to find q = q* which minimizes FPE(m*, q) and so on.

*ˆ and ˆ tt zz

t

q

jjtj

m

iititt

t

q

jjtj

m

iititt

xyzx

xyzy

2

*

1

**

1

**1202

111

1101

ˆ

ˆ

t

q

jjtj

m

iitit

t

q

jjtj

m

iitit

xyx

xyy

2

*

1

**

1

*02

111

01

nkn

mEESknmFPE

)(

)()()(

61

Granger causality

In regression analysis, it is usually assumed that the movements in the dependent variable are caused by movements in the independent variables, but the existence of a relationship proves neither the existence of causality nor its direction. In econometrics, Granger developed a special definition of causality which states that a variable x Granger causes y if prediction of the current value of y is enhanced by using past values of x (x → y).Returning to our example, given two stationary variables, say xt and yt , if we want to test if xt → yt , then the standard way is to test

H01 : 1 = 2 = …… = q = 0 vs.H11 : not Ho1

Using the associated F statistic However, given

that the model contains lagged dependent variables, the F statistic constructed is, strictly speaking, not F distributed under the null. Although most elementary textbooks suggest that F ~ F(q,n−k) , the correct statistic to use is Similarly, if we want to test if yt → xt , then we test

using the same procedure. If yt → xt and xt → yt , then bi-directional Granger causality exists between yt and xt . If only yt → xt or xt → yt , then causality is only uni-directional. Granger also points out that if a pair of variables are cointegrated, then there must exist Granger causality in a least one direction.

).p.32 see ,

)((

knESS

qESSESSF

UR

URR

.~ 2q

a

qF

212

**2

*102

not :

vs.0......:

o

m

HH

H

62

Are POP and XRATE cointegrated?

From previous lecture, we know that the New Zealand population over 1889 to 1991, POP, is I(1), and from the last tutorial exercise, we know that the New Zealand exchange rate (financial year average) over 1929 to 1991, XRATE, is also I(1). In this example, we aim to determine whether POP and XRATE are cointegrated. That is, is there a long run tendency for the two series to move together over time? Then the variables exactly offset each other to give a stationary linear combination. Apriori, I would not expect the two series to cointegrated.

We define the cointegrating regression as :

POPt = + bXRATEt + zt (1)

with the cointegrating residuals being

which should be I(O) if POP and XRATE are cointegrated. If not then zt (and so ) will be I(1). So, we want to test for the stationarity of zt using

ttt XRATEbPOPz ˆˆˆ

tz .ˆtz

63

1. CRDW testWe estimate (1) by OLS. The commands are as follows.data coint;infile C:\teaching\ar3316\nzpop2.dat ;input pop xrate;proc reg;model pop=xrate/dw;

Model: MODEL1Dependent Variable: POP


Source DFSum of

SquaresMean

Square F Value Prob > F

ModelErrorC Total

16162

38248.7165527819085.70727857334.423

38248.71655456050.58535

0.084 0.7731

Root MSEDep MeanC.V.

675.315172393.66508

28.21260

R-squareAdj R-sq

0.00140.0150

Variable DFParameter

EstimateStandard

ErrorT for H0:

Parameter = 0 Prob >T

INTERCEPXRATE

11

2207.808236204.813826

647.38115308707.22457751

3.4100.290

0.00120.7731

Parameter Estimates

64

Durbin-Watson D 0.003(For Number of Obs.) 631st Order Autocorrelation 0.964

65

Now, we wish to test

H0: DW = 0 ( is I(1), yt and xt are not cointegrated)

vs.

H1: DW > 0 ( is I(0), yt and xt are cointegrated)

Here, DW = 0.0032. Is this significantly different from zero?

We have 63 observations and so from the table of critical values, at the 10% significance level, d*(0.60, 0.32). Here DW is less than both the upper and lower bounds. This suggests that the residual series does possess a unit root. That is, zt is I(1) and POP and XRATE are not cointegrated.

tz

tz

66

2. CRADF test

Apply an augmented Dickey Fuller test to . That is, we estimate,

where p is selected to ensure that εt is approximately white noise.output out = out1 r=e;data coint2;set out1;le=lag(e);de=ele;proc reg;model de=le;output out=out2 r=e2;Run;

tz

t

p

jjtjtt zzz

11 ˆˆˆ (2)

67

Model: MODEL1Dependent Variable: DE


Source DFSum of

SquaresMean


ModelErrorC Total

16061

141.5573326172.3545526313.91188

141.55733436.20591

0.325 0.5710

Root MSERoot MSEC.V.

20.8855431.6312866.02813

R-squareAdj R-sq

0.00540.0112


EstimateStandard

ErrorT for H0:Parameter = 0 Prob >T

INTERCEPLE

11

31.6690430.002300

2.653294800.00403696

11.9360.570

0.00010.5710

Parameter Estimates

68

So, we first need to determine p as per our discussion on this issue for the integration topic. Following are the SAS commands and output with p = 0

69

proc arima data=out2;identify var=e2 nlag=10;run;data; ARIMA Procedure

Name of variable = E2.

Mean of working series = 103E 16Standard deviation = 20.54592Number of observations = 62

Autocorrelations

Lag Covariance Correlation -1 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 1

0 422.135 1.00000 * * * * * * * * * * * * * * * * * * * *

1 208.151 0.49309 * * * * * * * * * *

2 140.963 0.33393 * * * * * * *

3 44.684601 0.10585 * *

4 79.969036 0.18944 * * * *

5 65.670205 0.15557 * * *

6 63.614366 0.15070 * * *

7 99.981992 0.23685 * * * * *

8 123.477 0.29250 * * * * * *

9 114.144 0.27040 * * * * *

10 50.277969 0.11910 * *

marks two standard errors

70


Lag Correlation -1 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 1

1 0.49309 * * * * * * * * * *

2 0.11996 * *

3 0.13164 * * *

4 0.20137 * * * *

5 0.03595 *

6 0.00820

7 0.22379 * * * *

8 0.12530 * * *

9 0.00302

10 0.08356 * *

The SAS System

Some of the autocorrelations appear significant, So we now try p =1.

71

set coint2;lde=lag(de);proc reg;model de=le lde;output out=out3 r=e3;run;proc arima data=out3;identify var=e3 nlag=10;

Model: MODEL1Dependent Variable: DE


Source DFSum ofSquares

MeanSquare F Value Prob > F

ModelErrorC Total

25860

6504.4257819723.6598926228.08568

3252.21289340.06310

9.564 0.0003


18.4408031.7819258.02292

R-squareAdj R-sq

0.24800.2221


EstimateStandard


INTERCEPLELED

111

16.0352480.0002550.496475

4.331482990.003647360.11447506

3.7020.0704.337

0.00050.94440.0001

Parameter Estimates

72

ARIMA Procedure


Mean of working series = 345E 17Standard deviation = 17.98162Number of observations = 61

Autocorrelations


0 323.339 1.00000 * * * * * * * * * * * * * * * * * * * *

1 20.739251 0.06414 *

2 52.790556 0.16327 * * *

3 53.140142 0.16435 * * *

4 44.839909 0.13868 * * *

5 9.241376 0.02858 *

6 0.094331 0.00029

7 30.430160 0.09411 * *

8 48.580814 0.15025 * * *

9 55.844914 0.17271 * * *

10 2.255192 0.00697


73


Lag Correlation -1 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 1

1 0.06414 *

2 0.15981 * * *

3 0.14972 * * *

4 0.10475 * *

5 0.08903 * *

6 0.06162 *

7 0.12263 * *

8 0.18218 * * * *

9 0.13934 * * *

10 0.00064

74

The SAS System

This looks reasonable and so we’ll proceed using p = 1 (note that we only have critical values for p = 0 and p = 4). So, our CRADF regression is

We test H0 : = 0 vs. H0 : < 0 using the t statistic on . This t statistic is 0.0700. The appropriate 10% critical value, say t* for p = 0 is t* (3.28, 3.03) and for p = 4 is t* (2.9, 2.91). As our t statistic is positive, we cannot reject H0. That is, we believe that the residuals do have a unit root – that is, POP and XRATE are not cointegrated.

In this example, we are fortunate that our two tests indicate the same result. This is not always the cases.

t

p

jjtjtt zzz

11 ˆˆˆ (3)

75

As discussed in lectures, asymptotically, it makes no difference whether we consider

POPt = + bXRATEt + zt

or

XRATEt = * + b*POPt + zt*

However, it can matter in small samples. So we need to repeat the above exercises estimating the reverse cointegrating regression,

XRATEt = * + b*POPt + zt* (4)

and consider whether zt* is I(0) or I(1).

1. CRDW test

As before, we estimate (4) by OLS. The SAS commands and output are as follows:

76

data;set cont;proc reg;mode xrate=pop/dw;output out=out4 r=e4;run;

Model: MODEL1Dependent Variable: XRATE


Source DFSum ofSquares

MeanSquare F Value Prob > F

ModelErrorC Total

16162

0.001250.910550.91180

0.001250.01493

0.084 0.7731


0.122180.90744

13.46377

R-squareAdj R-sq

0.00140.0150


EstimateStandard


INTERCEPPOP

11

0.8913960.000006704

0.057507230.00002315

15.5010.290

0.00010.7731

Parameter Estimates

77

Durbin-Watson D 0.222(For Number of Obs.) 631st Order Autocorrelation 0.863

Here DW = 0.2223. As before, the 10% critical value d* (0.69, 0.32). DW is lower than either bound and so we cannot reject the hypothesis that DW = 0: XRATE and POP are not cointegrated.

78

2. CRADF test

As before, we first need to identify the degrees of augmentation required. The commands and output are as follows:

data coint3;set out4;le4=lag(e4);de4=e4le4;proc reg;model de4=le4;output out=out5 r=e5;run;proc arima data=out5;identify var=e5 nlag=10;run;

79

Model: MODEL1Dependent Variable: DE4


Source DFSum of

SquaresMean


ModelErrorC Total

16061

0.008880.192120.20101

0.008880.00320

2.775 0.1010


0.056590.00476

1188.64230

R-squareAdj R-sq

0.04420.0283


EstimateStandard


INTERCEPLE4

11

0.0044540.100806

0.007188910.06051752

0.6201.666

0.53790.1010

Parameter Estimates

80

ARIMA Procedure


Mean of working series = 1.18E 18Standard deviation = 0.055667Number of observations = 62

Autocorrelations


0 0.0030988 1.00000 * * * * * * * * * * * * * * * * * * * *

1 0.00048471 0.15642 * * *

2 0.0004054 0.13082 * * *

3 0.0001191 0.03844 *

4 0.00062939 0.20311 * * * *

5 0.00049964 0.16124 * * *

6 0.0004079 0.13162 * * *

7 0.000121 0.03903 *

8 0.000131 0.04228 *

9 0.00034508 0.11136 * *

10 0.0000561 0.01810


81

Inverse Autocorrelations

Lag Correlation -1 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 1

1 0.22531 * * * * *

2 0.10467 * *

3 0.00149

4 0.12426 * * * *

5 0.11126 * *

6 0.16614 * * *

7 0.10656 * *

8 0.13013 * * *

9 0.08022 * *

10 0.00020

82


Lag Correlation -1 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 1

1 0.15642 * * *

2 0.15918 * * *

3 0.01071

4 0.19711 * * * *

5 0.09343 * *

6 0.13644 * * *

7 0.05213 *

8 0.11440 * *

9 0.09270 * *

10 0.00023

83

The SAS System

This suggests that we do not require any augmentation. So, our CRADF regression is

We test H0 : = 0 vs. H1 : < 0 using the t statistic on which is 1.666. Here p = 0 so from the critical value tables we know that at the 10% significance level t* (3.28, 3.03). Our t-statistic is greater than either bound – we cannot reject H0. XRATE and POP are not cointegrated. Again, we are fortunate that we have no conflicting results.

* * *1ˆ ˆt t tz z (9)

84

SOME CRITICAL VALUES FOR SOME COINTEGRATION TESTS

Cointegrating regression is:

yt = + Xt + zt

where yt and Xt are I(1). z is I(0) if yt and Xt are contegrated.

Significance Level

Test Sample Size T 1% 5% 10%

(a) CRADF

(i) p = 0 50 4.32 3.67 3.28

100 4.07 3.37 3.03

200 4.00 3.37 3.02

(ii) p = 4 50 4.12 3.29 2.90

100 3.73 3.17 2.91

200 3.78 3.25 2.98

(b) CRDW 50 1.00 0.78 0.69

100 0.51 0.39 0.32

200 0.29 0.20 0.16

Source: Engle, R.F. & B.S. Yoo, 1987, Forecasting & Testing in Co-Integrated Systems, Journal of Econometrics, 35, 143-159.

Documents

Chapter 5 : Unit-root Testing and Cointegration Analysis